Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $a = \dfrac{n + 9}{3n - 18} \times \dfrac{3n^2 - 6n - 72}{-3n^2 - 36n - 81} $
Solution: First factor out any common factors. $a = \dfrac{n + 9}{3(n - 6)} \times \dfrac{3(n^2 - 2n - 24)}{-3(n^2 + 12n + 27)} $ Then factor the quadratic expressions. $a = \dfrac {n + 9} {3(n - 6)} \times \dfrac {3(n - 6)(n + 4)} {-3(n + 9)(n + 3)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac {(n + 9) \times 3(n - 6)(n + 4) } {3(n - 6) \times -3(n + 9)(n + 3) } $ $a = \dfrac {3(n - 6)(n + 4)(n + 9)} {-9(n + 9)(n + 3)(n - 6)} $ Notice that $(n + 9)$ and $(n - 6)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac {3(n - 6)(n + 4)\cancel{(n + 9)}} {-9\cancel{(n + 9)}(n + 3)(n - 6)} $ We are dividing by $n + 9$ , so $n + 9 \neq 0$ Therefore, $n \neq -9$ $a = \dfrac {3\cancel{(n - 6)}(n + 4)\cancel{(n + 9)}} {-9\cancel{(n + 9)}(n + 3)\cancel{(n - 6)}} $ We are dividing by $n - 6$ , so $n - 6 \neq 0$ Therefore, $n \neq 6$ $a = \dfrac {3(n + 4)} {-9(n + 3)} $ $ a = \dfrac{-(n + 4)}{3(n + 3)}; n \neq -9; n \neq 6 $